Chapter 14, Problem 11E

To determine

Whether the mean or the median would be better for the student if the three recorded grades were $99,88,95$ or if the grades were $99,70,77$.

Answer to Problem 11E

Solution:

The median would be better for the student.

Explanation of Solution

The given data is,

$\begin{array}{rll}\text{A}& =& 99,88,95\\ \text{B}& =& 99,70,77\end{array}$

The formula to calculate the mean of set of data is,

$\begin{array}{rll}\text{Mean}\left(\overline{x}\right)& =& \frac{x_1+x_2+\mathrm{...}+x_n}{n}\\ & =& \frac{{\displaystyle \sum _{i1}^n{x}_i}}{n}\end{array}$

Here, $x_i$ is the $i^{\text{th}}$ data value and $n$ is the number of data values in the data set.

Substitute $\text{99},88,95$ for $x_1,x_2,x_3$ respectively in the formula.

Here number of data in the given data set is $3$, substitute $3$ for $n$ in the formula.

$\begin{array}{rll}\text{Mean}\left(\overline{x}\right)& =& \frac{x_1+x_2+\mathrm{...}+x_n}{n}\\ & =& \frac{99+88+95}{3}\\ & =& 94\end{array}$

Mean of A is $94$.

Substitute $99,70,77$ for $x_1,x_2,x_3$ respectively in the formula.

$\begin{array}{rll}\text{Mean}\left(\overline{x}\right)& =& \frac{x_1+x_2+\mathrm{...}+x_n}{n}\\ & =& \frac{99+70+77}{3}\\ & =& 82\end{array}$

Mean of B is $82$.

The formula to calculate the median of set of ordered data is,

$\text{Median}=\{\begin{array}{l}{\left(\frac{n+1}{2}\right)}^{\text{th}}\text{value},n=\text{odd}\\ \frac{1}{2}\left[{\left(\frac{n}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{n+1}{2}\right)}^{\text{th}}\text{value}\right],n=\text{even}\end{array}$

Here $n$ is the number of data values in the data set.

Arrange the given data in increasing order.

$\begin{array}{rll}\text{A}& =& 88,95,99\\ \text{B}& =& 70,77,99\end{array}$

Number of values in data set is $3$, it is even.

Substitute $3$ for $n$ in the formula for odd number of data set A.

$\begin{array}{rll}\text{Median}& =& {\left(\frac{n+1}{2}\right)}^{\text{th}}\text{value}\\ & =& {\left(\frac{3+1}{2}\right)}^{\text{th}}\text{value}\\ & =& {\text{2}}^{\text{nd}}\text{value}\\ & =& \text{95}\end{array}$

The median of A is $95$.

Substitute $3$ for $n$ in the formula for odd number of data set B.

$\begin{array}{rll}\text{Median}& =& {\left(\frac{n+1}{2}\right)}^{\text{th}}\text{value}\\ & =& {\left(\frac{3+1}{2}\right)}^{\text{th}}\text{value}\\ & =& {\text{2}}^{\text{nd}}\text{value}\\ & =& \text{77}\end{array}$

The median of B is $77$.

Since, mean deviation about mean is zero, whereas, mean deviation about median is minimum. Hence, median would be better for the student.

MATLAB Code:

A = [99 88 95];

B = [99 70 77];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

Save the MATLAB script with the name, Chapter14_54793_14_11E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Therefore, the required program is stated above.